Weil cohomologies seem to be "natural" and useful cohomology theories. Wikipedia lists Betti, De Rham, 𝓁adic étale, and crystalline cohomologies as examples of Weil cohomology. Do we have more of them? and is it plausible to classify all or some Weil cohomologies? Or more generally, classify these really good Grothendieck sites?
2 Answers
If we believe in the standard conjectures (or something similar) so that the category of motives is Tannakian, then a Weil cohomology theory is just a fibre functor and as such twists of each other (that does not quite give the multiplicative structure however) which gives, more or less, a classification. Note that your question on Grothendieck sites is only vaguely related, the whole point of the notion of Weil cohomology theory is that there is no site in sight (pun intended).
Addendum: This is a little bit simple minded as the requirements for a Weil cohomology theory to be a fibre functor go beyond the standard conjectures I think. Also one would need to define motives using cohomological equivalence. However, I think the statements I made are philosophically OK and one cannot hope to get anything in the way of a more precise classification.

$\begingroup$ Torsten, I'm probably reading too much into your choice of words, but I'd disagree that "the whole point of the notion of Weil cohomology theory is that there is no site in sight". I might say that since the definition of Weil cohomology theory doesn't involve a choice of site, it would be particularly satisfying if, when the universal Weil cohomology theory is discovered, it can be constructed without any particular choice of site. I might even disagree with that, because all definitions of scheme (or algebraic space) essentially involve some choice of site. $\endgroup$– JBorgerJun 4, 2010 at 6:08

$\begingroup$ My comment was more referring to what I believe to be Grothendieck's motivation. Finding a natural way to interpret Weil cohomology theories as the cohomology with respect to some topology (and some sheaf in it) would be an a posteriori fact (and a wonderful one at that). $\endgroup$ Jun 4, 2010 at 6:13

$\begingroup$ would any of the listed four functors serve as a kind of special functors? or they are just "one of them"? By the way, do you know more examples of Weil cohomologies? Thank you. $\endgroup$– naturaJun 4, 2010 at 6:35

3$\begingroup$ An interesting (yet curious) example is an ultraproduct of etale coomology with $\mathbb Z/\ell$coefficients over all $\ell$ different from the characteristic. It was used by Gabber to show that $\ell$adic cohomology is torision free for all but a fini for a finite number of $\ell$ (and for a fixed smooth and projective variety). $\endgroup$ Jun 4, 2010 at 12:50

1$\begingroup$ I can't remember if Weil cohomology theories are allowed to take values in categories more general than vector spaces over a field of characteristic 0, but modulo this issue, big de RhamWitt cohomology is another example. Unfortunately, nothing about it has appeared in the literature yet, but you won't have to wait long. $\endgroup$– JBorgerJun 5, 2010 at 1:26
The answer to this question is yes. Roughly speaking: make a category Weil(A) of Weil cohomologies with values in a rigid tensor abelian category A and let Weil() be the resulting 2functor. Then Weil() is 2representable by the universal Weil cohomology. See B. Kahn's lecture Universal Weil cohomology.
Universal cohomologies are already explained here in an answer to What is a cohomology theory (seriously)?.